Q.131−1002−23 Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
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The Inverse Matrix Method
Many problems reduce to a system of linear equations, for example
2x+3yx− y=8=−1.
The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.
Writing the system as AX=B
Collect the coefficients, the unknowns and the constants:
A=(213−1),X=(xy),B=(8−1),
so the whole system becomes AX=B.
The idea
For numbers, ax=b gives x=a−1b provided a=0. The same works for matrices: if A is invertible, multiply AX=B on the left by A−1:
A−1(AX)=A−1B⇒IX=A−1B⇒X=A−1B.
X=A−1B
Multiplying on the left matters — matrix products do not commute, so BA−1 would be wrong.
When it works
The inverse A−1 exists only when detA=0, so:
- detA=0: the system is consistent with the unique solution X=A−1B.
- detA=0: no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.
Worked steps
For the system above, detA=(2)(−1)−(3)(1)=−5=0, and
A−1=−51(−1−1−32).
Then …
Concept: Inverse of a matrix via elementary row operations (Gauss-Jordan method).
We augment the given matrix with the identity and row-reduce until the left side becomes I.
Step 1: Write the augmented matrix [A∣I].
131−1002−23100010001
Step 2: Eliminate below the first pivot.
R2→R2−3R1, R3→R3−R1:
100−1312−811−3−1010001
Step 3: Swap R2 and R3 to get a pivot in row 2, then eliminate.
R2↔R3:
100−11321−81−1−3001010
R3→R3−3R2:
100−11021−111−1000101−3
Step 4: Back-substitute to obtain I on the left.
R3→−111R3:
100−1102111−1000−11101113
R2→R2−R3, R1→R1−2R3: …
Here det(A)=11=0, so the inverse exists. By the adjoint method, A−1=1110−11031−1283.
For A=131−1002−23, use A−1=detA1adj(A).
1. Determinant (expand along column 2, which has two zeros):
det(A)=(−1)(−1)1+231−23=(−1)(−1)(9+2)=11=0
2. Cofactors Cij=(−1)i+jMij:
C11=00−23=0,C12=−31−23=−11,C13=3100=0
C21=−−1023=3,C22=1123=1,C23=−11−10=−1 …
Method: Finding the Inverse of a 3×3 Matrix Using the Adjoint
The standard adjoint-method procedure for a general square matrix whose inverse is required.
Steps
Step 1: Compute ∣A∣ and confirm it is nonzero
Expand along the row/column with the most zeros. If ∣A∣=0, the inverse exists; if ∣A∣=0, stop — the matrix is singular and has no inverse.
Step 2: Compute all nine cofactors Cij=(−1)i+jMij
Delete the relevant row and column for each entry to form its 2×2 minor, then apply the sign.
Step 3: Form the adjoint as the transpose of the cofactor matrix
adj(A)=C11C12C13C21C22C23C31C32C33. …
Common Mistakes
Mistake 1: Forgetting the transpose step when forming the adjoint from the cofactor matrix
Why it's wrong: the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself — using the untransposed version silently swaps off-diagonal entries and produces a wrong inverse. Correct approach: explicitly write (adjA)ij=Cji (swap row/column indices) when assembling the adjoint from the computed cofactors.
Mistake 2: Making a sign error in one of the nine cofactors, especially when a minor itself contains a negative entry …
- AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL6 marksQ.Using matrix method solve the following system of linear equations: x−y+z=4, 2x+y−3z=0, x+y+z=2. OR Using elementary transformation find the inverse of the following matrix: A=1−32305−2−50.
›Reveal solutionSolution
Solve AX=B via X=A−1B using the adjoint method; for the OR, row-reduce [A∣I] to [I∣A−1].
Matrix method: x−y+z=4, 2x+y−3z=0, x+y+z=2
A=121−1111−31, X=xyz, B=402
detA=1(1+3)−(−1)(2+3)+1(2−1)=4+5+1=10=0, so a unique solution exists.
Cofactors: C11=4, C12=−5, C13=1, C21=2, C22=0, C23=−2, C31=2, C32=5, C33=3
adj(A)=4−5120−2253, so A−1=1014−5120−2253
X=A−1B=1014(4)+2(0)+2(2)−5(4)+0(0)+5(2)1(4)−2(0)+3(2)=10120−1010=2−11
Check: 2−(−1)+1=4✓, 4+(−1)−3=0✓, 2−1+1=2✓.
So x=2, y=−1, z=1.
OR: inverse of A=1−32305−2−50 by elementary row operations
Write [A∣I] and reduce:
R2→R2+3R1, R3→R3−2R1 gives rows (1,3,−2∣1,0,0), (0,9,−11∣3,1,0), (0,−1,4∣−2,0,1).
R3→−R3: (0,1,−4∣2,0,−1); swap R2,R3: rows become (1,3,−2∣1,0,0), (0,1,−4∣2,0,−1), (0,9,−11∣3,1,0).
…
- AHSEC Higher Secondary (HS) Final Examination 2018Set ANNUAL6 marksQ.Solve the following system of equations by matrix method: 3x−2y+3z=8, 2x+y−z=1, 4x−3y+2z=4.
›Reveal solutionSolution
Writing AX=B with ∣A∣=−17e0, the system has the unique solution x=1, y=2, z=3.
Write the system as AX=B with A=324−21−33−12, X=xyz, B=814.
Determinant: ∣A∣=3(1⋅2−(−1)(−3))−(−2)(2⋅2−(−1)⋅4)+3(2⋅(−3)−1⋅4)=3(−1)+2(8)+3(−10)=−3+16−30=−17.
Since ∣A∣=−17e0, A−1 exists and X=A−1B gives a unique solution. Solving (by X=A−1B, or by elimination): from the second equation y=1−2x+z; substituting into the first gives 7x+z=10, and into the third gives 17x=17.
Hence x=1, then z=10−7(1)=3, and y=1−2(1)+3=2.
Check: 3(1)−2(2)+3(3)=8 ✓; 2(1)+2−3=1 ✓; 4(1)−3(2)+2(3)=4 ✓.
…
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